(New page: Now we know that <br> <math> x(t)</math> ⇒ <math>X(\omega)</math><br> Now suppose the input signal was multiplied by a cosine wave then the fourier transform of the wave would look a...)
 
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Now suppose the input signal was multiplied by a cosine wave then the fourier transform of the wave would look as follows
 
Now suppose the input signal was multiplied by a cosine wave then the fourier transform of the wave would look as follows
  
<math>x(t)*cos(t)</math> &rArr; <math>\frac{\pi}{j}[\delta(\omega - \pi) - \delta(\omega + \pi)]</math>.
+
 
 +
==<math>x(t)*cos(t)</math> &rArr; <math>\frac{\pi}{j}[\delta(\omega - \pi) - \delta(\omega + \pi)]</math>. ==

Revision as of 10:07, 24 October 2008

Now we know that
$ x(t) $$ X(\omega) $

Now suppose the input signal was multiplied by a cosine wave then the fourier transform of the wave would look as follows


$ x(t)*cos(t) $$ \frac{\pi}{j}[\delta(\omega - \pi) - \delta(\omega + \pi)] $.

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang